Here $\sigma(T):=\{c\in\Bbb{C}|\ T-cI\text{ is not invertible}\}$. I know that the above result holds for compact,self adjoint operators.
I can write $T=A+iB$ where $A=\frac{T+T^*}{2}$ and $B=\frac{T-T^*}{2i}$. Here $A,B$ is compact and self adjoint operator with $AB=BA$ (Since $T$ is normal).
I think anyhow I need to use the result for compact, self-adjoint operator (on $A,B$ possibly) to prove the similar result for $T$ (compact, normal).
But I am not getting any idea how to proceed. Can anyone help me in this regard? Thanks for help in advance.
Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then $$ \lambda B v = B \lambda v = BA v = AB v, $$ so $Bv$ is an eigenvector of $A$ with eigenvalue $\lambda$ too. Hence $B \ker (A - \lambda I) \subset \ker (A - \lambda I)$. And the same holds for $B$.
Denote $(\alpha_i)_{1\le i \le N_A}$ and $(\beta_j) _{1\le j\le N_B}$ the sequences of non-zero eigenvalues of $A$ and $B$ respectively.
Therefore, for each eigenvalue $\alpha_i$ of $A$ and for each eigenvalue $\beta_j$ of $B$ $\ker (A - \alpha_i I) \cap \ker (B - \beta_j I)$ is an invariant subspace for both $A$ and $B$. Let $(\varphi^{ij}_k)_{1\le k \le N_{ij}} $ be an orthonormal basis of this subspace (or just one zero vector if the intersection is zero). Hence $$ A = \sum_{i = 1}^{N_A}\sum_{j = 1}^{N_B}\sum_{k=1}^{N_{ij}}\alpha_{i} \langle \varphi_{k}^{ij}, \cdot \rangle \varphi_{k}^{ij}, $$ $$ B = \sum_{i = 1}^{N_A}\sum_{j = 1}^{N_B}\sum_{k=1}^{N_{ij}}\beta_{j} \langle \varphi_{k}^{ij}, \cdot \rangle \varphi_{k}^{ij}. $$ (I used the Hilbert—Schmidt decomposition here).
Therefore, $$ T = \sum_{i = 1}^{N_A}\sum_{j = 1}^{N_B}\sum_{k=1}^{N_{ij}}(\alpha_{i}+i\beta_j) \langle \varphi_{k}^{ij}, \cdot \rangle \varphi_{k}^{ij}. $$